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In article <1162257513.017058.157220@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
barr <barr@xxxxxxxxx> wrote:
> It is well known and easy to prove that square roots of complex numbers
> can be calculated using real square roots. That is you can solve the
> equation (x + iy)^2 = a + ib for x and y using only real square roots.
> My question is, is it possible to calculate complex cube roots using
> only real cube and, if necessay, square roots. I tried, but the
> equations I got seemed intractible.
>
The famous "casus irriducibilis" !
Google it.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
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